5

A side-rational triangle stands for a triangle with each side rational.

We know, by cosine theorem and computing the area of the triangle, we can get that each angle of the triangle is of rational $\sin$ and $\cos$. Then consider ant height of one side, we have $$S=h(h\cot \alpha+h\cot \beta)/2$$ Then it suffices to show that $$t-\frac{1}{t}+T-\frac{1}{T}\notin \mathbb{Q}^2\quad t,T\in \mathbb{Q}$$ But i am stuck on it.

Cubic Bear
  • 1,582
  • 1
    You could use Heron's formula to work on it. – Ghartal Aug 18 '17 at 14:16
  • 1
    Set $p:=\frac{a+b+c}{2}$, where $a,b,c$ are the sides of the triangle. Then, by Erone's formula, we have equivalently $$ \sqrt{p(p-a)(p-b)(p-d)}=1, $$ i.e., $$ (a+b+c)(a+b-c)(a-b+c)(-a+b+c)=16. $$ – Paolo Leonetti Aug 18 '17 at 14:17
  • Set $a:=A/D, b:=B/D, c:=C/D$ and $P:=A+B+C$, then we have to solve in $\mathbf{N}$ $$ P(P-2A)(P-2B)(P-2C)=(2D)^4. $$ All the factors have the same parity. So $P$ is even, hence we obtain $$ \frac{P}{2}\left(\frac{P}{2}-A\right)\left(\frac{P}{2}-B\right)\left(\frac{P}{2}-C\right)=D^4. $$ – Paolo Leonetti Aug 18 '17 at 14:23
  • Of course, there is no rational right triangle with area $1$, since $n=1$ is not congruent - see here. – Dietrich Burde Aug 18 '17 at 14:49

2 Answers2

4

A triangle with sidelengths (3/2, 5/3, 17/6) works. The upper coordinate is at {45/34, 12/17}. Here's a picture of it with a unit square. Any heronian triangle with a square area can be used, here are two more: (3, 25, 26)/6, (17, 113, 120)/30. For more, see the Sascha Kurz heronian triangle list.

area 1 triangle

A few more: (28,1345,1371)/84, (41,357,370)/84, (104,657,697)/180, (17,4904,4905)/204, (289,2313,2600)/204, (73,1274,1299)/210, (255,353,392)/210, (40,8749,8787)/234, (305,424,567)/252, (337,441,680)/252, (130,2057,2169)/264, (520,641,1089)/330, (1009,3088,4095)/336, (113,3137,3150)/420, (245,1443,1448)/420, (800,1241,2009)/420, (56,9273,9305)/462, (539,890,1233)/462, (585,746,847)/462, (696,865,1183)/546, (2809,4395,7202)/546, (314,2555,2619)/630, (833,1017,1066)/630, (4201,4375,8574)/630, (1000,2057,2993)/660, (377,4570,4879)/714, (544,2329,2535)/780, (1409,6596,7995)/780, (801,1825,2176)/840, (1131,1285,1904)/840, (401,4097,4290)/858, (841,1898,2307)/870, (1617,2425,3944)/924, (1445,1476,2281)/1020, (939,2548,2785)/1092, (1241,6884,8075)/1140

Ed Pegg
  • 20,955
0

Well, the formula itself Geronova (Heronian) triangle.

$$S_g=\sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}$$

If: $p,s,k,t$ -integers asked us. Then the solutions are.

$$a=(pt+ks)(k^2+t^2)ps$$

$$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$

$$c=(pt+ks)(p^2+s^2)kt$$

$$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$

individ
  • 4,301
  • Do you have a reference for this? – John Bentin Aug 18 '17 at 15:37
  • @JohnBentin no. This formula I found. – individ Aug 18 '17 at 15:39
  • @JohnBentin I found one similar and quadrangle. https://math.stackexchange.com/questions/979009/how-to-find-all-cyclic-quadrilaterals-with-integer-sides/980656#980656 – individ Aug 18 '17 at 15:46
  • What restrictions are there on $p,t,k,$ and $s$ ? For example, it's obvious that none of them can be zero. Also, it's easy to see that all four—but not one or three—of them could be negative. – John Bentin Aug 18 '17 at 16:40
  • @JohnBentin this is not so important. It is possible for the total share is always reduced. – individ Aug 18 '17 at 16:46